Did you know that not all molecules are reactive? Only certain types of molecules will want to react in a mechanism. Let’s dig a little deeper into this...
Stability and reactivity generally have an inverse relationship. If a molecule is unstable in some way, it will want to react! Here are the 4 signs we can look for that determine reactivity:
Concept: How to tell if a molecule will be reactive or not.6m
Hey guys! So now we’re going to get into the part that makes this chapter really fun and that’s the mechanism. Alright? So just so you guys know, I’m going to teach this section more in depth than your professor did for this chapter. Why is that? Well, it’s because I realized that mechanisms are such a big part of organic chemistry that if we can learn them as early as possible it’s going to help you guys so much more later on. Okay? You don’t want to be learning about mechanisms for the first time in Chapter 7 or Chapter 8. That would be organic suicide. Alright? Because you’re going to be very confused and you’re going to have to pick up the pace a lot.
So what we want to do is we’re want to just use the concepts of acids and bases to introduce what the mechanism is, alright? So we’re going to talk about this in two parts. The first part is I want to talk about the reactivity of atoms. Why do they even react? And the second part is I want to show you how they react with electron movement. So let’s go ahead and begin.
So just so you guys know, the currency of organic chemistry, and I’ve said this before, is stability. Okay. Stability. Every atom wants to become more stable. Every molecule wants to become more stable. Alright? And it turns out that stability and reactivity, you hear these words all the time, actually have an inverse relationship. Inverse. What I mean by that is opposite. That means if stability goes up, reactivity is generally going to go down. If reactivity is up, that means stability generally goes down. They’re opposites of each other. Okay? So how can we tell if a molecule is reactive and if a molecule is unreactive? Alright. Again, this is not a section that is usually covered by your textbook. I have just built this from years of teaching Orgo and realizing that hey, there’s actually ways we can predict this. Okay.
So there’s actually four really common way to know that something is going to be reactive. You guys ready? And you actually know three of these four already. Okay? The first one is formal charges. Do you guys remember those? Okay. Formal charges, what does that mean? Formal charges mean that an atom is not at its ideal bonding preferences, right? It has too many or too few valence electrons. Okay? Remember the sticks and the dots. So what that means is that it’s going to try to do everything in its power to try to go back to its bonding preferences. So that means it has a very big reason to want to react to something else. If it can get rid of a few electrons maybe it can go back to the state that it wants to be in. Does that make sense? So that’s the first one. If you see a formal charge you know that this is going to be reactive.
The second one is net dipoles. Okay? So net dipoles have to do with the fact that I have asymmetrical dipoles that are not perfectly cancelling out and what that means is that I’m going to have partial charges in different places. Another way that I could say this is partial charges. Okay? So remember that a partial charge is the result of a net dipole. If I have a net dipole going to the right that means that I would have a partial negative on one side and a partial positive on the other. Okay? So what I’m basically doing here is I’m saying that if you have formal charges—a full charge—that’s very reactive. But even if you have a partial charge that’s also very reactive.
Let’s now talk about two other things. The two other things are π bonds. Okay? What we’re going to find out is that π bonds, remember that we went over this in the first chapter? π bonds are not as strong as σ bonds, right? They’re actually much weaker in terms of the amount of energy that is saved. So what that means is that π bonds are really a good source of electrons. Okay? And they’re going to be pretty easy to break. Remember, σ bonds are almost impossible to break many times. But π bonds, they’re not as stable so they’re easier to break. So π bonds are going to also be a lot reactive. I know I said the word “break” a lot of times.
And finally the fourth one is something you don’t know yet and that is steric effects. Okay? Steric effects is something that we’re talking about more when we talk about basically cycloalkanes and we’re going to talk about a bunch of other stuff a little bit later. But an example of steric effects, I just want to tell you it would be something like ring strain. So imagine if I have a three-member ring and these bond angles are very, very, very tight. Usually those bond angles would want to be like 109.5° but instead they are like 60°. That’s an example of ring strain. That means that these bond angles are not stable because of the fact that they’re, you know, they’re just not at their ideal position. Alright? And that’s also going to make molecules reactive. Now, do you need to deal with these right now? No. But I just want to be comprehensive and let you know like those are the four major things that make organic molecules reactive. Cool?
So what I want to do is move on to an exercise and I want you guys to just look at these eight different compounds. And just go ahead and look at the first box. The first box is literally a checkbox. So I want you to either check off that it is reactive or put an X on it if it is not reactive. So I want you to maybe go through A through D, first four, and I want you guys to look at those first four and see if you can find any of those four reasons why it would be reactive. If you can’t then that means it’s not going to be reactive.
Example: Are any of the following 4 molecules reactive?2m
Alright, so A through D. Let’s just go ahead and just start one at a time. Does A have a formal charge? No, it doesn’t. It’s fine. Does it have a partial charge? Actually it does. Okay? It doesn’t have a double bond and it’s not straight. Remember that I told you the steric effects are usually just when you have like these small rings. So you don’t really have to worry about it. Okay? But A is going to be reactive because of the fact that A has a net dipole. So if I were going to draw the dipole here I would draw it towards the chlorine. That means I’ll have a negative and a positive. Okay? And what that means is that this is going to be reactive because I have partial charges. Is that cool so far? Cool. We don’t know how it’s going to react but we just know it’s reactive.
Let’s look at the second one. The second one was easy. It has a positive charge. That’s a formal charge. So this one is definitely going to be reactive. Okay?
Let’s look at this next one. This one has a ton of dipole moments. It actually has four dipole moments. So you would expect this to be super reactive, right? It’s just going to like blow up, right? Actually, no. This turns out to be inert. Do you guys know what inert means? It means it doesn’t react at all. It doesn’t react with anything. And the reason is that it doesn’t have a net dipole. Remember that you can have dipole moments but if they’re perfectly balanced it’s not going to be a net dipole. So in this case this one actually would not react with anything because it does not have a new dipole. This is a tetrahedral and it’s perfectly balanced. Alright? Then lastly, so I’m just going to put an X there. Not reactive.
And then the last one is a double bond. And remember that I said double bonds are easy to break and that they are good sources of electrons. So this is going to be reactive as well. So basically I have three reactive molecules, I said four, three reactive molecules in different ways and then one reactive molecule.
These reactive trends ARE in order of strength (i.e. a formal charge will typically be more reactive than a dipole).
Example: Are any of the following 4 molecules reactive?2m
Again, we are literally just matching these molecules to the 4 patterns discussed above.
Concept: How to tell if charged molecules will react as nucleophiles or electrophiles.3m
But now how are they going to react? That’s the next question. Okay? What are they going to react as? And it turns out that we can basically characterize all reactive organic molecules—this is important—into two different subtypes. Okay? And those two subtypes are basically negatively charged species and positively charged species.
Okay, so a negatively charged species, I know that you have heard this before, but I’m just going to remind you what they are. A negatively charged species is known as a nucleophile. Okay? And the way that I’m going to, in this course, abbreviate nucleophile is I’m often going to write Nu(-). Okay? And that stands for the fact that I have a negatively charged atom, okay, a negatively charged atom or a negatively charged molecule. And the important thing here is that I’m generalizing because it’s going to react in about the same way no matter what type of nucleophile it is. As long as it’s a nucleophile it’s going to react in certain types of reactions. Alright. Then the positively charged ones are going to be known as electrophiles. Okay? Electrophiles, I’m going to go ahead and summarize them using E(+), meaning that it’s a positively charged substance.
And I also do want to talk about the names a little bit. What does nucleophile mean and electrophile? Well if it’s negatively charged that means that what it wants is positive in order to balance out, right? So nucleophile literally means it’s a nucleus lover or it’s a proton lover. Okay? So a nucleophile is going to want positive charges like protons in order to balance. Same thing with an electrophile. An electrophile has a positive charge so it’s going to want electrons to bond with or to react with in order to balance that positive with a negative. Alright? So that’s how we think about it. But if you just want, just think that the nucleophile is the negative and the electrophile is the positive. Alright?
So now we look at all of these reagents. Okay, I’m giving you eight. And what I want you guys to do is figure out which of these are going to react more as electrophiles and which of them are going to react more as electrophiles. For a few of these that’s easy. For a few of these we can automatically tell. Okay. So let me go ahead and tell you guys which ones are the easy ones to identify. Okay? B is easy to identify. E is easy to identify. And actually D is easy to identify. Okay, so for these, already just without knowing the rule I’m going to teach you, you should be able to tell me if these are going to react as nucleophiles or electrophiles.
Now that we know how to determine if molecules are reactive, we still don’t know HOW they will react! There are two major subtypes of reactivity that we’ll often use in Orgo 1 and 2:
Note that a molecule doesn’t require a negative charge to be a nucleophile, but it needs to have similar properties (i.e. a source of electrons).
That said, try to identify if the following three molecules are nucleophilic or electrophilic.
Example: Are the following 3 reactive molecules nucleophiles or electrophiles?2m
Concept: How to tell if uncharged molecules will react as nucleophiles or electrophiles.3m
So now we just solved three out of eight. Three out of eight were really easy. Now for the next ones they are actually be tougher and ready to know a new rule. Because notice all these other ones have both positive and negative regions. So how do I know if it’s a nucleophile or an electrophile? And that actually is one of the biggest questions that I get when we’re at this section. “Johnny, how do I know it’s a nucleophile? How do I know it’s an electrophile?” And it turns out we have this rule that we can use to figure out if it’s going to react more like a nucleophile or react more like an electrophile. Is that cool? So I’m going to teach you that rule.
And what that rule is this: the side of the dipole that has the highest bonding preferences can be used to predict the nucleo-felicity or electro-felicity. So what that means is that I agree with you. There are two charges on all of these molecules. On the five molecules that are left there are two charges, there’s a positive and there’s a negative. Okay. Let’s look at A. But what I’m going to do is I’m going to say which of the sides is the one that can make the most bonds? So for example, the positive charge is on a carbon. How many bonds can carbon make? Four. Okay. The negative charge is on a chlorine. How many bonds can chlorine make? One. Okay, remember according to bonding preferences. And I told you, you’re going to keep using bonding preferences all semester. So is this going to react more like a nucleophile or more like an electrophile? And the way that I look is I just basically look at number four. I look at the highest bonding preference and I look at that charge. That’s the one I pay attention to because that’s the one with the highest bonding preference. So the answer is that this is going to be what? This is actually going to be a really good electrophile. Why? Because the atom that has the highest bonding preferences is the one that is positively charged. Isn’t that cool? And that’s what you’re going to do with all of these.
So what I want you guys to do is we’ve already figured out B, what about C? Is C a good electrophile or nucleophile? I forgot to say this earlier, I should have pointed this out. Is C a nucleophile or an electrophile? This is actually one of the easy ones too, I forgot. So there was actually four easy ones and four hard ones. C is easy. Okay, so I’m going to highlight this as well. The answer is that it is none of them. Okay? Because I just told you, it’s not even reactive at all so if it’s not reactive at all, how can it be a proton-lover or an electron-lover? It’s a nothing lover. It just loves itself. It’s a narcissist, okay? So it’s just going to be inert. It’s not going to react to anything so I wouldn’t consider it as an electrophile or a nucleophile. Alright?
So what that means is that really this has become a lot easier for you guys. All you need to do is identify F, G and H and tell me what you guys think in terms of if they are going to be electrophiles or nucleophiles, what do you guys think? So go ahead and pause the video now and use that rule to determine these last three.
So that wasn’t so hard, but those were the easy cases. What if you have nucleophilic AND electrophilic regions on the same molecule? Is it possible to determine how it will react? Yes it is!
Rule: The side of the dipole with the highest bonding preference (the atom that wants to make the most bonds) will determine how the molecule reacts.
Example: Are the following 3 reactive molecules nucleophiles or electrophiles?3m
Now we understand which molecules will want to react, and we are getting better at determining If they are nucleophiles or electrophiles, but how to they actually attack other molecules?
Concept: Learning the rules of electron movement5m
So that first one was pretty fun, right? We just learned how to tell the difference between a non-reactive molecule which is basically just inert or a reactive molecule which is actually going to try to make bonds with other molecules. Okay? On top of that, then we also have what type of reactivity they’ll have—if they’re going to react more like an electrophile or nucleophile. Well, now what I want to do is actually learn, I want to teach you guys how to start drawing out these mechanisms. What that means is we’re actually going to start drawing electron movement and you guys are probably going to be the first people in your class to draw as good mechanisms as we are going to do now. But I’m not concerned. I know you guys are smart and I think you guys are going to like this part. So let’s go ahead and get started.
So reactive molecules are the ones that are going to try to do something about their reactivity. What that means is that if you’re a non-reactive molecule, you’re inert. You don’t care. But if you’re reactive you’re going to try to share electrons with another molecule in order to become more stable. Remember that we talked about that in the first chapter—how we form molecular orbitals in order to reach a lower energy state. Okay?
So it turns out that we’re going to use curved arrows to indicate the direction that these electrons are going. And in Organic Chemistry that’s actually a huge part of this field. A huge part of this field is just figuring out where these electrons are moving to. They start here and they move over here and then what do they do and then they move over here. And those are called mechanisms. Okay? So I’m just going to tell you guys right now: a huge part of this course is going to be you drawing arrows and figuring out where these electrons move from atom to atom to atom. Okay? So I just want to teach you guys or remind you guys about the way electrons move and it turns out that these rules that I’m going to teach you are the same that we learned for resonance structures. Why? Because it turns out that resonance structures are the same thing in terms of that we’re removing electrons from one atom to another. So let’s go ahead and get started.
So remember that arrows are always going to move from regions of high electron density to low electron density. Okay? So what that means is that we always want to start from the part of the molecule that has the most electrons and move to the part of the molecule that doesn’t have a whole lot. This is the same thing that we did with resonance structures. Okay?
But back when we talked about resonance structures we didn’t know about nucleophiles and electrophiles. So now that I’ve taught you what a nucleophile is and what an electrophile is, which one do you think is always going to be the one that starts off the arrow? Because remember, we always start from the high area. Okay? So do we start from the nucleophile or from the electrophile? And the answer is that we’re going to start off from the nucleophile. Okay? Because the nucleophile is the one with the negative charge, right? So we’re always going to have nucleophiles that attack electrophiles. Okay? And a huge part of organic chemistry can be summed up at saying nucleophiles attack electrophiles. Honestly, when it comes down to it, you can probably explain 50% of the reactions just talking like that. Okay? So that’s a very important point. Okay?
So another thing is that each attacking arrow is going to represent two electrons that are being shared. So every time that I draw one of these arrows that means that I’m sharing two electrons with another atom. And what that means is that after the reaction has taken place I’m going to replace whatever that arrow was and I’m going to replace it with a new σ bond. Okay? Why is that? Because remember that a σ bond equals two electrons being shared. Okay? So what that means is that an arrow is really just a fancy way of drawing a σ bond in motion. That means I’m creating a new σ bond. Does that make sense so far? I know it’s a lot of words. So let’s just go ahead and get started.
So what I want you guys to do for these reactions is I know that you don’t even know what these molecules are, really. But we do understand nucleophiles and we do understand electrophiles. And we also understand how electrons move now. We kind of know. So what I want you guys first of all is to determine the initial direction of electron movement by drawing the very first arrow of each mechanism. Even though we don’t know what’s going on here, it’s okay because you can just follow the general rule of nucleophile and electrophile.
Reactive molecules share electrons to become more stable. Arrows are used to show which direction they are going.
Summary: Molecules with lots of electrons will attack (draw an arrow to) molecules with a positive charge. Let’s get drawing!
Example: Draw the first arrow of the following mechanism.1m
So in this case I have a negative charge and I have a positive charge. Where do you think my arrow is going to start from? My arrow is going to start form the negative charge. Because remember, that you always start from the nuclear fault or the area of higher density. So what that means is that my arrow starts from the negative. Where do you think it’s going to go to? Well, it’s going to go to the area of the least electrons. That means it’s actually going to try to attack that carbon right there because what’s going on is that the negative has too many electrons and it’s saying, “Hey, I want to go ahead and link up to a positive charge in order to neutral out and to balance out.” Does that make sense? So the first arrow of my mechanism is negative to positive. Does that make sense so far? That’s very similar to what we did in resonance structures except that now I have a full negative and a full positive. Okay?
Example: Draw the first arrow of the following mechanism2m
Example: Draw the first arrow of the following mechanism2m
So now we know how to make bonds. Do we ever have to break bonds? How do we know if we do or we don’t?
Concept: Why we need to break bonds sometimes.2m
Now what I want to talk about is what we just did is what I call bond making. All of this was forms of bond-making where all I am—basically all these arrows indicate the sharing of two electrons, right? And I said that after the reaction you’re going to replace that arrow with a σ bond. Okay? So that means all of these arrows that I just drew are eventually going to become little sticks that I draw off of each of these atoms. So that’s going to be a stick. That’s going to be a stick. And that’s going to be a stick. I hope that you understand that what I’m saying is that arrow is just a really long way of drawing a single bond that’s going to be formed. Okay?
Now what I want to talk about is bond breaking. Okay? So bond-breaking is sometimes required, okay, in mechanisms. But it’s only required when it’s needed to preserve octets. So what that means is that just like resonance structures, remember that sometimes we would want to make a bond but that bond would break the octet of a certain atom. So then what would we do to fix that? We would also break a bond. We would also make a bond and then break a bond in order to preserve the octet of that atom. So in the same way we’re going to want to break bonds in order to preserve octets of atoms that are getting attacked. Okay?
Bond breaking is sometimes required in mechanisms, but only when it is necessary to preserve octets.
Concept: The two ways to break bonds.2m
Now there's actually two ways to break bonds there and I just want to teach as this really quick, there's Heterolytic cleavage which Heterolytic cleavage means that you get ions, OK? You get like a cation on one side and you get an anion on the other, OK? Heterolytic, Hetero means different so just means that you get different amounts of electrons on both, OK? Notice that for a Heterolytic cleavage I use a full arrow, OK? Well then, the other option is that I have Homolytic cleavage, Homolytic cleavage means that each of the atoms once the bond breaks gets the same amount of electrons so each of them get a radical, OK? And Homolytic cleavage is done with half arrows, notice that the head of that arrow is a little bit smaller it's like only big on one side, OK? I know that my head's right in the way of radicals but you're going to get radicals, OK? So, what I want us to know is that for the most for this entire chapter we're really just going to be dealing with Heterolytic cleavage we're not going to get into Homolytic cleavage until we talk about radical chemistry which is coming up in a few chapters so you don't have to worry about it yet, Cool?
Out of these two different ways, we will stick to heterolytic cleavage for the foreseeable future (we won’t discuss radicals for a few more chapters).
Example: Identify if the following reaction requires a bond to be broken. Draw the products.2m
So what I want you guys to do now is we already talked about which bonds we would be made. What I want you guys to do now is figure out which bonds would also be broken in order to not violate octets. Basically what I want you guys to do is finish these mechanisms on your own. Okay? I know that sounds rough but let’s go ahead and look at A. Okay? So for A, in my first step I went ahead and I made a bond to this carbon. Okay? Now that I’ve made a bond to this carbon am I’m violating its octet? The answer is no, I’m not. Because check it out. This carbon had how many, I’m sorry, had how many electrons before? Two, four, six. How many total electrons can carbon have? Eight. Eight, okay? So if I add two more is that going to break the octet? No. So that means that in this mechanism there’s going to be no bond breaking. That mechanism is over. So my final product would look like this. O, so I just redrew the original. Now for the way that original where that curved arrow was, now I’m just going to draw a single bond. That single bond represents the two electrons that are being shared. And now I’m going to draw that connected to what was on the other side. So that’s going to be attached to a C with three H’s. Isn’t that interesting? That’s my final product. So basically all I did was I made a bond but then I never violated the octet so that’s my final product. Okay? And once again that new blue bond comes from the arrow. Okay? The arrow turned into a bond.
Example: Identify if the following reaction requires a bond to be broken. Draw the products.3m
Example: Identify if the following reaction requires a bond to be broken. Draw the products.3m
Provide a mechanism for the following reaction. Be sure to include all intermediates, formal charges and arrows depicting electron movement.
For each of the following one step transformations show the movement of electrons by using the standard curved arrow notation. Show all formal charges for each structure on the left and right of the reaction arrows.
Follow the flow of electron indicated by the curved arrows predict the products of the reaction.
Provide the product(s) for the following mechanistic steps based on the curved arrows show.
Draw the curved arrow that shows the electron flow and indicate the pattern (proton transfer, nucleophilic attack, carbocation rearragement, or loss of leaving group), for each of the following reaction steps.
Draw the curved arrow mechanism and products for the following reaction.
Draw curved arrows on the reactant side of the reaction arrow to show how electron pairs move in the formation of the indicated products.
Predict the product(s) of the following reactions:
Supply curved arrows for the following reactions.
Consider the structural formula of acetone given below.
a) Using Lewis structures for each reactant and each product and using curved arrows appropriately to show the flow of electrons in the reactants, give a chemical equation for the reaction between hydroxide ion (HO− ) and acetone in which HO− acts as a Lewis base and acetone acts as a Lewis acid. IMPORTANT: For this part a of problem 8, the Lewis base that you give cannot also be a Brønsted-Lowry base; the Lewis acid that you give cannot also be a Brønsted-Lowry acid.
b) Using Lewis structures for each reactant and each product and using curved arrows appropriately to show the flow of electrons in the reactants, give a chemical equation for the reaction between hydroxide ion (HO− ) and acetone in which HO− acts as a Brønsted-Lowry base dehyde acts as a Brønsted-Lowry acid. Do not give the same answer that you gave in part a of this problem.
Recall that for a Lewis structure, you must show all atoms, all bonding valence electrons, all nonbonding valence electrons, and all nonzero formal charges.
Given the reaction below, please predict the curved arrow mechanism.
What pattern of curved arrow pushing is the second step of this reaction?
a) Proton transfer
c) Loss of leaving group
d) Nucleophilic attack
For each step of the following mechanism:
a) Draw curved arrows
b) Identify arrow-pushing patterns
The following reaction has three mechanic steps. Draw all curved arrows necessary to complete the mechanism.
Which one represents a bond heterolysis?
The following is a generic depiction of a reaction using the curved arrow formalism.
Which of these statements is not correct for this reaction?
a. Electrons move from C to B.
b. In the products, A would have a positive charge.
c. In the products, a bond forms between C and B.
d. Electrons move from B to A.
Use the reaction below to answer the following questions:
a. Draw an arrow(s) to indicate the movement of electrons.
b. Draw reasonable products for the following acid base reactions.